- #1

- 281

- 60

## Main Question or Discussion Point

Would this ##f'(x)>g'(x)\,\forall x\in [a,\infty)\text{ and }f,\,g\underset{\infty}{\to}0\Rightarrow \lim_{x\to\infty}f(x)/g(x)=0## hold if ##\frac{f(x)}{g(x)}\neq c##?

- B
- Thread starter archaic
- Start date

- #1

- 281

- 60

Would this ##f'(x)>g'(x)\,\forall x\in [a,\infty)\text{ and }f,\,g\underset{\infty}{\to}0\Rightarrow \lim_{x\to\infty}f(x)/g(x)=0## hold if ##\frac{f(x)}{g(x)}\neq c##?

- #2

Math_QED

Science Advisor

Homework Helper

2019 Award

- 1,349

- 484

What is ##a## ? What is ##c##? Where are ##f,g## defined?

- #3

RPinPA

Science Advisor

Homework Helper

- 552

- 316

This is a little cryptic. Here's what I think you are saying:Would this ##f'(x)>g'(x)\,\forall x\in [a,\infty)\text{ and }f,\,g\underset{\infty}{\to}0\Rightarrow \lim_{x\to\infty}f(x)/g(x)=0## hold if ##\frac{f(x)}{g(x)}\neq c##?

"There is a theorem that if for some ##a \in \mathbb R, f'(x)>g'(x)\,\forall x\in [a,\infty)## and ##f,\,g\underset{\infty}{\to}0## (Does this mean ##\lim_{x\to\infty}f, g = 0##?) then ##\lim_{x\to\infty}f(x)/g(x)=0##. Does the theorem hold if ##\frac{f(x)}{g(x)}\neq c##?"

Or are you asking if there is such a theorem?

- #4

WWGD

Science Advisor

Gold Member

2019 Award

- 5,122

- 2,321

If I understood correctly, ##1/x ,1/(x+1)## are counters.

- #5

- 281

- 60

What is ##a## ? What is ##c##? Where are ##f,g## defined?

What I want to say is, if there're two functions ##f(x)## and ##g(x)## who are going to ##0## as ##x\to\infty##, and such that after a certain ##x\,(=a)## we have ##f'(x)>g'(x)## then ##\frac{f(x)}{g(x)} \underset{\infty}{\to} 0## provided that ##f(x)\neq cg(x)## where ##c\in\mathbb{R}##, elsewise the limit would equal ##c##.This is a little cryptic. Here's what I think you are saying:

"There is a theorem that if for some ##a \in \mathbb R, f'(x)>g'(x)\,\forall x\in [a,\infty)## and ##f,\,g\underset{\infty}{\to}0## (Does this mean ##\lim_{x\to\infty}f, g = 0##?) then ##\lim_{x\to\infty}f(x)/g(x)=0##. Does the theorem hold if ##\frac{f(x)}{g(x)}\neq c##?"

Or are you asking if there is such a theorem?

I realised some holes in this through math.stackexchange. I was initially thinking of two decreasing functions, i.e both positive.

However, since what I want to say is if ##f## goes to ##0## faster than ##g## then ..., the correct way to translate this in math is ##|f'(x)|<|g'(x)|## (provided that both ##f## and ##g## are going to ##0##).

- #6

- 281

- 60

No, they don't come under this since they have the same rate of change.If I understood correctly, ##1/x ,1/(x+1)## are counters.

- #7

WWGD

Science Advisor

Gold Member

2019 Award

- 5,122

- 2,321

Respective derivatives are ##-1/x^2 >-1/(x+1)^2##No, they don't come under this since they have the same rate of change.

- #8

- 281

- 60

Right sorry. I think, then, that I can say something about the convergence of the limit given the assumptions? And then it follows that it is ##0## if the functions are not equivalent at ##\infty##?Respective derivatives are ##-1/x^2 >-1/(x+1)^2##

- #9

- 11,266

- 4,448

I can't see any reason why this would be true.What I want to say is, if there're two functions ##f(x)## and ##g(x)## who are going to ##0## as ##x\to\infty##, and such that after a certain ##x\,(=a)## we have ##f'(x)>g'(x)## then ##\frac{f(x)}{g(x)} \underset{\infty}{\to} 0## provided that ##f(x)\neq cg(x)## where ##c\in\mathbb{R}##, elsewise the limit would equal ##c##.

I realised some holes in this through math.stackexchange. I was initially thinking of two decreasing functions, i.e both positive.

However, since what I want to say is if ##f## goes to ##0## faster than ##g## then ..., the correct way to translate this in math is ##|f'(x)|<|g'(x)|## (provided that both ##f## and ##g## are going to ##0##).

- #10

- 281

- 60

$$N=n-m>0,\,\,\pm\frac{10^{-n}}{10^{-m}}=\pm10^{-n+m}=\pm10^{-N}=\pm\frac{1}{e^{N\ln10}}\underset{N\to \infty}{\to}0$$I can't see any reason why this would be true.

By ##10^{k}## I just wanted to represent a kind of manual evaluation of a limit where the nominator is going to zero faster than the denominator. Maybe I should use ##O(10^{-n})##? Not sure, I need to study big-O notation.

- #11

- 11,266

- 4,448

I can't see the relationship between that and what you posted before.$$N=n-m>0,\,\,\pm\frac{10^{-n}}{10^{-m}}=\pm10^{-n+m}=\pm10^{-N}=\pm\frac{1}{e^{N\ln10}}\underset{N\to \infty}{\to}0$$

By $10^{k}$ I just wanted to represent a kind of manual evaluation of a limit where the nominator is going to zero faster than the denominator. Maybe I should use ##O(10^{-n})##? Not sure, I need to study big-O notation.

- #12

- 281

- 60

Well, if ##|f'(x)|<|g'(x)|## and they are both going to ##0##, it means that ##f## is going to ##0## faster, and so the limit should be ##0##.I can't see the relationship between that and what you posted before.

- #13

- 11,266

- 4,448

That relationship means that ##g(x)## is changing more quickly than ##f(x)##.Well, if ##|f'(x)|<|g'(x)|## and they are both going to ##0##, it means that ##f## is going to ##0## faster, and so the limit should be ##0##.

You've got that the wrong way round.

- #14

- 11,266

- 4,448

In any case, you already have the counterexample: ##f(x) = \frac 1 2 g(x)##.Well, if ##|f'(x)|<|g'(x)|## and they are both going to ##0##, it means that ##f## is going to ##0## faster, and so the limit should be ##0##.

Even if you exclude that precise counterexample, you only need two functions where ##f(x) \approx \frac 1 2 g(x)##. For example:

##f(x) = (\frac 1 2 + \frac{\sin x}{x})g(x)##

- #15

- 281

- 60

If ##f## is going faster to ##0## than ##g##, then ##|df|## would be less than ##|dg|##, because it is already "at zero" so there isn't much to change.That relationship means that ##g(x)## is changing more quickly than ##f(x)##.

You've got that the wrong way round.

Right, see post #8.In any case, you already have the counterexample: ##f(x) = \frac 1 2 g(x)##.

Even if you exclude that precise counterexample, you only need two functions where ##f(x) \approx \frac 1 2 g(x)##. For example:

##f(x) = (\frac 1 2 + \frac{\sin x}{x})g(x)##

through the different counter-examples given, it is clear that this is not true. however, given the premises, it seems that the limit always converges (maybe), and in the special case that ##f\underset{\infty}{\not\sim}g##, it is zero (maybe). This is just an intuitive exercise, I just feel like this might work out.

- #16

- 11,266

- 4,448

Not that either. Just amend my last function:however, given the premises, it seems that the limit always converges (maybe), and in the special case that ##f\underset{\infty}{\not\sim}g##, it is zero (maybe). This is just an intuitive exercise, I just feel like this might work out.

##f(x) = (\frac 1 2 + \frac{\sin x}{10})g(x)##

Then the limit ##f(x)/g(x) = \frac 1 2 + \frac{\sin x}{10}## is undefined.

- #17

- 281

- 60

wopsNot that either. Just amend my last function:

##f(x) = (\frac 1 2 + \frac{\sin x}{10})g(x)##

Then the limit ##f(x)/g(x) = \frac 1 2 + \frac{\sin x}{10}## is undefined.

- #18

- 281

- 60

It doesn't satisfy ##|f'(x)|<|g'(x)|## so it doesn't come under this.Not that either. Just amend my last function:

##f(x) = (\frac 1 2 + \frac{\sin x}{10})g(x)##

Then the limit ##f(x)/g(x) = \frac 1 2 + \frac{\sin x}{10}## is undefined.

- #19

- 11,266

- 4,448

Seriously, you ought to think about how I created these counterexamples. They only really involved some elementary geometrical thinking.wops

The only one we don't have is where ##\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = +\infty##.

Any ideas?

- #20

- 11,266

- 4,448

With ##g(x) = e^{-x}## it all works out.It doesn't satisfy ##|f'(x)|<|g'(x)|## so it doesn't come under this.

- #21

- 281

- 60

you mean given the premises?Seriously, you ought to think about how I created these counterexamples. They only really involved some elementary geometrical thinking.

The only one we don't have is where ##\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = +\infty##.

Any ideas?

- #22

- 11,266

- 4,448

Yes.you mean given the premises?

- #23

- 11,266

- 4,448

Maybe this is the result that could be proved? If ##|f'(x)| < |g'(x)|##, then ##f(x)## is changing less than ##g(x)##, so ##f(x)/g(x)## cannot become arbitrarily large. So:you mean given the premises?

##\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} \ne +\infty##?

- #24

- 281

- 60

I thought you meant that there was an example where that limit goes to infinity...Maybe this is the result that could be proved? If ##|f'(x)| < |g'(x)|##, then ##f(x)## is changing less than ##g(x)##, so ##f(x)/g(x)## cannot become arbitrarily large. So:

##\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} \ne +\infty##?

But yes, it seems reasonable that given those premises the limit can't be arbitrarily large.

- #25

- 11,266

- 4,448

... looking for counterexamples is good way to see what might be true.I thought you meant that there was an example where that limit goes to infinity...

- Last Post

- Replies
- 1

- Views
- 2K

- Replies
- 10

- Views
- 1K

- Last Post

- Replies
- 14

- Views
- 925

- Last Post

- Replies
- 9

- Views
- 2K

- Replies
- 2

- Views
- 2K

- Replies
- 5

- Views
- 5K

- Last Post

- Replies
- 4

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 1K